Question

An 8-lb weight is placed at the end of a coil spring suspended from the ceiling. After coming to rest in its equilibrium position, the weight is set into vertical motion and the period of the resulting motion is 4 sec. After a time this motion is stopped, and the 8 -lb weight is replaced by another weight. After this other weight has come to rest in its equilibrium position, it is set into vertical motion. If the period of this new motion is \(6 \mathrm{sec}\), how heavy is the second weight?

Short answer

The second weight is \(22.5 \) pounds.

Step-by-step solution

1. Write down the formula for the period of oscillation

We will be using the formula for the period of oscillation of a mass-spring system, which is given as: \( T = 2\pi\sqrt{\frac{m}{k}} \), where \(T\) is the period, \(m\) is the mass, and \(k\) is the spring constant.

2. Set up an equation with the given information for the first weight

We know that for the first weight, its mass is 8 lb (converted to slugs, where 1 slug = 32.2 lb) and its period is 4 seconds. So, let's plug those values into the equation: \(4 = 2\pi\sqrt{\frac{m_1}{k}} \) Here, \(m_1 = \frac{8}{32.2} \) slugs and \(k\) is the spring constant.

3. Solve the equation for the spring constant 'k'

We want to find the spring constant, so we need to solve the equation for 'k': \(4 = 2\pi\sqrt{\frac{m_1}{k}} \) Square both sides to get rid of the square root: \(16 = 4\pi^2\frac{m_1}{k} \) Now, solve for 'k': \(k = \frac{4\pi^2 m_1}{16} = \frac{\pi^2 m_1}{4} \)

4. Find the value of the spring constant 'k'

We can now substitute the value of \(m_1\) to find the value of the spring constant 'k': \(k = \frac{\pi^2 \frac{8}{32.2}}{4} = \frac{2\pi^2}{32.2} \) slugs/ft

5. Set up equation for second weight

Now, we will set up the equation for the second weight with the given period, 6 seconds: \(6 = 2\pi\sqrt{\frac{m_2}{k}} \) We now need to solve for the second mass, \(m_2\).

6. Solve for the second mass 'm_2'

First, square both sides to get rid of the square root: \(36 = 4\pi^2\frac{m_2}{k} \) Now, solve for 'm_2': \(m_2 = \frac{36k}{4\pi^2} \)

7. Substitute the value of 'k' to find the mass 'm_2'

We can now substitute the value of 'k' into our equation for 'm_2': \(m_2 = \frac{36 \times \frac{2\pi^2}{32.2}}{4\pi^2} = \frac{9}{8} \) slugs Now, convert the mass back to pounds: \(m_2 = \frac{9}{8} \times 32.2 = 22.5 \) lb

8. State the final answer

The second weight is 22.5 pounds.

Key concepts

Mass-Spring System

A mass-spring system is a classic model used in physics to study oscillatory motion. When a mass is attached to a spring, it can be set into motion by displacing the mass from its equilibrium position. The system then exhibits a regular, repeating pattern known as oscillation.
An essential aspect of the mass-spring system is its ability to oscillate without any external influence once set into motion. This can be observed when you stretch or compress the spring and then let it go. The mass will move up and down, passing through its original position repeatedly.
Understanding the dynamics of this system is crucial for grasping how mechanical vibrations work, be it in simple toys or complex machinery.

• Motion type: Harmonic
• Energy exchange: Between potential and kinetic energy

Period of Oscillation

The period of oscillation refers to the time it takes for the mass-spring system to complete one full cycle of motion. It is a crucial parameter in understanding the behavior of oscillatory systems. In the context of a mass-spring system, the period is defined by the formula:
\[ T = 2\pi\sqrt{\frac{m}{k}} \]
In this equation, \(T\) is the period, \(m\) is the mass attached to the spring, and \(k\) is the spring constant. This formula shows us that the period depends on both the mass and the stiffness of the spring.
If the mass increases, the period typically lengthens, meaning it takes more time for the system to complete one cycle. Similarly, a stiffer spring (higher \(k\)) leads to a shorter period. Grasping this concept is vital for predicting how changing parameters can affect the motion of the system.

• Unit: seconds (s)
• Inversely related to frequency

Spring Constant

The spring constant, denoted as \( k \), is a measure of a spring's stiffness. It quantifies the force required to extend or compress the spring by a unit length. In the formula for the period of oscillation,
\[ T = 2\pi\sqrt{\frac{m}{k}} \]
\(k\) plays a vital role. The greater the value of \(k\), the stiffer the spring, which generally results in a faster oscillation and a shorter period.
When we talk about the spring constant, it is expressed in units of force per unit length, such as pounds per foot (lb/ft) or newtons per meter (N/m). This is central to the mechanics of the system, as it determines the system's response to masses added to the spring.

• Reflects spring stiffness
• Higher \(k\) means faster oscillations

Physics Application

Physics applications of the mass-spring system extend far beyond classroom exercises. This simple model has practical implications in various fields, from engineering to biology.
In engineering, understanding how springs and masses interact is crucial for designing buildings, vehicles, and machinery in a way that minimizes harmful vibrations. This knowledge is applied in seismic retrofitting of buildings, designing vehicles for smoother rides, or even in crafting musical instruments.
In biology, similar principles help to understand phenomena such as the human ear's response to sound vibrations. The ear's inner workings resemble a series of mass-spring systems, translating mechanical vibrations into signals which the brain interprets as sound.

• Engineering: vibration analysis
• Biology: understanding auditory systems